Prove that the set {X: XA = AX} is a subspace of M 2,2

  • #1

Homework Statement


A is a 2 x 2 matrix. Prove that the set W = {X: XA = AX} is a subspace of M2,2


Homework Equations





The Attempt at a Solution


I have already proven non-emptiness and vector addition.

Non-emptiness:

Code:
W must be non-empty because the identity matrix I is an element of W.
IA = AI
A = A
Vector addition:

Code:
Let X, Y be elements of W such that XA = AX, YA = AY. Add the equations together.
(X+Y)A = A(X+Y)
XA+YA = AX+AY

Hence, X+Y is an element of W.
I have no idea what to do with scalar multiplication. This is my current attempt:

Code:
Let X be an element of W, and c be a scalar.
cXA = AcX
(cX)A = A(cX)
I also tried this:

Code:
Let X be an element of W, and c be a scalar.
XA = AX
c(XA) = c(AX)
I don't feel like either of these attempts proves anything.

I don't understand how to prove that cX is actually in W. This doesn't make any sense to me. How do I prove that?
 

Answers and Replies

  • #2
MathematicalPhysicist
Gold Member
4,294
203
It's too trivial.
You have X in W, thus XA=AX, thus cXA=A(cX) meaning cX in W.
Or in other word you've proved it already.
 
  • #3
It's too trivial.
You have X in W, thus XA=AX, thus cXA=A(cX) meaning cX in W.
Or in other word you've proved it already.
But how do I know that there isn't a scalar that would change X in such a way that (cX)A ≠ A(cX)?

Edit: I guess I get what you're saying now, but it still confuses me how I can just say that they're equal with no real evidence.
 
  • #4
MathematicalPhysicist
Gold Member
4,294
203
They are equal cause you have XA=AX, and then you multiply this equation by c.
I said it's trivial.
 
  • #5
MathematicalPhysicist
Gold Member
4,294
203
Remember that your scalar is some number, real/complex or something else.
It's not a vector (in which case multiplication of a matrix with a vector will not yield you a matrix of the same size but a vector, and thus it wouldn't be a vector space.
 
  • #6
Yeah, I understand it now. This is the first class I've had to write my own proofs in, so I'm still getting used to it. Thanks again for the help.
 

Related Threads on Prove that the set {X: XA = AX} is a subspace of M 2,2

Replies
4
Views
463
Replies
12
Views
3K
Replies
2
Views
1K
Replies
5
Views
3K
Replies
8
Views
970
Replies
6
Views
1K
Replies
44
Views
16K
Replies
6
Views
8K
Replies
9
Views
1K
Top